Let the complex numbers \(\alpha\) and \(\frac{1}{\bar{\alpha}}\) lie on the circles \(\left|z-z_{0}\right|^{2}=4\) and \(\left|z-z_{0}\right|^{2}=16\) respectively, where \(z_{0}=1+i\). Then, the value of \(100|\alpha|^{2}\) is

Complex Numbers — JEE Main · 2024 · 27 Jan 2024 · Shift 2 Question with Solution

Let the complex numbers

α

and

\frac{1}{α ˉ}

lie on the circles

∣ z - z_{0} ∣^{2} = 4

and

∣ z - z_{0} ∣^{2} = 16

respectively, where

z_{0} = 1 + i

. Then, the value of

100∣ α ∣^{2}

Answer: 20

Let

α = x + i y

z_{0} = 1 + i

, so

∣ z_{0} ∣^{2} = 2

α

lies on

∣ z - z_{0} ∣^{2} = 4

∣ α ∣^{2} - α \overset{z}{ˉ}_{0} - \overset{α}{ˉ} z_{0} + ∣ z_{0} ∣^{2} = 4 \Rightarrow ∣ α ∣^{2} - (α \overset{z}{ˉ}_{0} + \overset{α}{ˉ} z_{0}) = 2. (1)

\frac{1}{α ˉ}

lies on

∣ z - z_{0} ∣^{2} = 16

(\frac{1}{α ˉ} - z_{0}) (\frac{1}{α} - \overset{z}{ˉ}_{0}) = 16.

Expanding:

\frac{1}{∣ α ∣ ^{2}} - \frac{z ˉ _{0}}{α ˉ} - \frac{z _{0}}{α} + ∣ z_{0} ∣^{2} = 16

. Multiplying through by

∣ α ∣^{2}

1 - (α \overset{z}{ˉ}_{0} + \overset{α}{ˉ} z_{0}) + ∣ z_{0} ∣^{2} ∣ α ∣^{2} = 16∣ α ∣^{2} \Rightarrow 1 - (α \overset{z}{ˉ}_{0} + \overset{α}{ˉ} z_{0}) = 14∣ α ∣^{2} . (2)

From (1),

α \overset{z}{ˉ}_{0} + \overset{α}{ˉ} z_{0} = ∣ α ∣^{2} - 2

. Substituting into (2):

1 - (∣ α ∣^{2} - 2) = 14∣ α ∣^{2} \Rightarrow 3 - ∣ α ∣^{2} = 14∣ α ∣^{2} \Rightarrow 15∣ α ∣^{2} = 3 \Rightarrow ∣ α ∣^{2} = \frac{1}{5} .

100∣ α ∣^{2} = 20.

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Integer answer